\section{可立则立\quad{}能曲必曲}
\setcounter{tuxxx}{2}%图编号从3开始
{\heiti 题解}：``可立则立''，是指被对方打吃的三路子，在可立的情况下，就要立一个，虽然不能逃走，但是多送一子反而能取得便宜。如图一中黑1的立。\par
\begin{psgopartialboard*}{(1,1)(10,6)}
	%1
	\markpos{A}{f}{2}

	\stone{black}{d}{3}
	\stone{black}{d}{4}	
	\stone{black}{f}{4}
	\stone{black}{f}{5}
	\stone{black}{g}{3}

	\stone{white}{e}{3}
	\stone{white}{f}{3}
	\stone{white}{g}{4}
	\stone{white}{h}{3}
	\stone{white}{k}{4}

	\setcounter{gomove}{0}%步数编号从1开始
	\move{g}{2}%1
	\move{h}{2}%2

\end{psgopartialboard*}\par
``能曲必曲''，是``可立则立''的第二步，它的范围自然要窄些，如图二中黑3的下法。但图一中黑就不宜在A位曲了。\par
\begin{psgopartialboard*}{(1,1)(10,6)}
	%2
	\stone{black}{d}{5}
	\stone{black}{e}{5}	
	\stone{black}{f}{4}
	\stone{black}{f}{5}
	\stone{black}{g}{3}

	\stone{white}{d}{3}
	\stone{white}{e}{4}
	\stone{white}{f}{3}
	\stone{white}{g}{4}
	\stone{white}{h}{3}

	\setcounter{gomove}{0}%步数编号从1开始
	\move{g}{2}%1
	\move{f}{2}%2
	\move{h}{2}%3
	\move{j}{3}%4

\end{psgopartialboard*}\par
立和曲是一种战术手段，这里都是弃子法的一种。它们的意义在于：在可以多送一子的情况下，多送一子使棋留有余味，并可以通过收气得利。\par
\begin{psgopartialboard*}{(1,1)(10,6)}
	%3
	\markpos{A}{e}{4}

	\stone{black}{d}{3}
	\stone{black}{d}{4}	
	\stone{black}{f}{4}
	\stone{black}{f}{5}
	\stone{black}{g}{3}
	
	\stone{white}{e}{3}
	\stone{white}{f}{3}
	\stone{white}{g}{4}
	\stone{white}{h}{3}
	\stone{white}{k}{4}

	\setcounter{gomove}{0}%步数编号从1开始
	\move{g}{5}%1
	\move{g}{2}%2

\end{psgopartialboard*}\par
\tx ：黑1不立就打，白2一手提净，白棋一点毛病也没有了。黑反生出A位缺陷。这是初学者易犯的错误。\par
\begin{psgopartialboard*}{(1,1)(10,6)}
	%4	
	\stone{black}{d}{3}
	\stone{black}{d}{4}	
	\stone{black}{f}{4}
	\stone{black}{f}{5}
	\stone{black}{g}{3}
	
	\stone{white}{e}{3}
	\stone{white}{f}{3}
	\stone{white}{g}{4}
	\stone{white}{h}{3}
	\stone{white}{k}{4}
	
	\stone[\marktr]{black}{g}{3}

	\stone[\marktr]{white}{k}{4}

	\setcounter{gomove}{0}%步数编号从1开始
	\move{g}{2}%1
	\move{h}{2}%2
	\move{g}{5}%3
	\move{h}{4}%4
	\move{e}{4}%5
	\move{f}{2}%6
	\move{e}{2}%7
	\move{g}{1}%8

\end{psgopartialboard*}\par
\tx ：黑1立，得弃子要领，白2后，生出各种余味。如图黑借弃子先手走厚实，和前图比较，得失是不难明了的。当然白棋也厚实了，但由于白本已有\stone[\marktr]{white}一子，子力偏于重复，其实正因为这点，黑\stone[\marktr]{black}断，再1位立弃子才有实战意义。不过黑1立作为战术手段来讲，无论如何，总是积极的。\par
\begin{psgopartialboard*}{(1,1)(10,6)}
	%5
	\markpos{A}{d}{2}

	\stone{black}{d}{5}
	\stone{black}{e}{5}	
	\stone{black}{f}{4}
	\stone{black}{f}{5}
	\stone{black}{g}{2}
	\stone{black}{g}{3}

	\stone{white}{d}{3}
	\stone{white}{e}{4}
	\stone{white}{f}{2}
	\stone{white}{f}{3}
	\stone{white}{g}{4}
	\stone{white}{h}{3}

	\setcounter{gomove}{0}%步数编号从1开始
	\move{c}{3}%1
	\move{h}{2}%2

\end{psgopartialboard*}\par
\tx ：黑1直接托，白2打后，缺少变化，黑A位扳的价值也打了对折。\par
\begin{psgopartialboard*}{(1,1)(10,6)}
	%6
	\markpos{A}{e}{2}

	\stone{black}{d}{5}
	\stone{black}{e}{5}	
	\stone{black}{f}{4}
	\stone{black}{f}{5}
	\stone{black}{g}{2}
	\stone{black}{g}{3}
	
	\stone{white}{d}{3}
	\stone{white}{e}{4}
	\stone{white}{f}{2}
	\stone{white}{f}{3}
	\stone{white}{g}{4}
	\stone{white}{h}{3}

	\setcounter{gomove}{0}%步数编号从1开始
	\move{h}{2}%1
	\move{j}{3}%2
	\move{c}{3}%3
	\move{j}{2}%4
	\move{g}{5}%5
	\move{h}{4}%6
	\move{d}{2}%7

\end{psgopartialboard*}\par
\tx ：黑1曲后，再3位托是正确的下法，走到黑7时，白虽仍可脱先，但黑留有A位先手吃两子的便宜。黑右边弃子虽然损失些，总的说来是便宜的。\par
\begin{psgopartialboard*}{(1,1)(10,6)}
	%7
	\stone{black}{d}{5}
	\stone{black}{e}{5}	
	\stone{black}{f}{4}
	\stone{black}{f}{5}
	\stone{black}{g}{2}
	\stone{black}{g}{3}
	
	\stone{white}{d}{3}
	\stone{white}{e}{4}
	\stone{white}{f}{2}
	\stone{white}{f}{3}
	\stone{white}{g}{4}
	\stone{white}{h}{3}
	
	\setcounter{gomove}{0}%步数编号从1开始
	\move{h}{2}%1
	\move{j}{3}%2
	\move{h}{4}%3
	\move{g}{5}%4
	\move{j}{4}%5
	\move{j}{2}%6
	\move{g}{6}%7

\end{psgopartialboard*}\par
在黑征子有利或其它特殊场合，图中黑3有如图七征吃白子或从6位外逃的手段，当然黑1曲已不是弃子。为了加深对本图例的理解，了解这些变化也是必要的。\par
\begin{psgopartialboard*}{(1,1)(10,6)}
	%8
	\markpos{A}{h}{4}
	\markpos{B}{j}{2}
	\markpos{C}{g}{5}
	
	\stone{black}{d}{5}
	\stone{black}{f}{4}	
	\stone{black}{f}{5}
	\stone{black}{g}{2}
	\stone{black}{g}{3}

	\stone{white}{d}{3}
	\stone{white}{f}{2}
	\stone{white}{f}{3}
	\stone{white}{g}{4}
	\stone{white}{h}{3}

	\setcounter{gomove}{0}%步数编号从1开始
	\move{h}{2}%1
	\move{j}{3}%2

\end{psgopartialboard*}\par
\setcounter{tuxxx}{7}%图编号从8开始
\tx ：``能曲必曲''，作为弃子，在实战中经常可以适用。如图，在角上没有托之类的直接便宜时，黑1曲，往往就是必要的，因为走出这个形来，总是要配合全局战斗的。白2后，如果黑在附近下子，这里白怎样补呢？如果A位粘，黑1和白2交换已明显便宜；如白在B位补，这样与被白直接在1位打相比，黑虽多死一子，却有了A位和C位打的种种利用。\par
\begin{psgopartialboard*}{(1,1)(8,8)}
	%9	
	\stone{black}{c}{4}
	\stone{black}{d}{4}	
	\stone{black}{d}{6}
	\stone{black}{e}{5}
	\stone{black}{e}{6}
	\stone{black}{g}{3}

	\stone{white}{d}{5}
	\stone{white}{d}{8}
	\stone{white}{e}{3}
	\stone{white}{e}{4}
	\stone{white}{f}{5}
	\stone{white}{f}{6}
	\stone{white}{g}{4}
	\stone{white}{h}{3}

\end{psgopartialboard*}\par
\tx （习题）：这是一间低夹定式后，黑棋连连脱先走出的棋形，如现在还该白走，当怎样下呢？\par
\begin{psgopartialboard*}{(1,1)(8,8)}
	%10
	\stone{black}{c}{4}
	\stone{black}{d}{4}	
	\stone{black}{d}{6}
	\stone{black}{e}{5}
	\stone{black}{e}{6}
	\stone{black}{g}{3}
	
	\stone{white}{d}{5}
	\stone{white}{d}{8}
	\stone{white}{e}{3}
	\stone{white}{e}{4}
	\stone{white}{f}{5}
	\stone{white}{f}{6}
	\stone{white}{g}{4}
	\stone{white}{h}{3}

	\stone[\marktr]{white}{d}{5}

	\setcounter{gomove}{0}%步数编号从1开始
	\pass*
	\move{c}{6}%1
	\move{c}{7}%2
	\move{d}{7}%3
	\move{c}{5}%4
	\move{b}{6}%5
	\move{b}{7}%6
	\move{e}{7}%7
	%\move{d}{14}%8=\stone[\marktr]{white}
	\pass
	\move{b}{8}%9
	\move{b}{5}%10
	\move{c}{8}%11
	\move{a}{6}%12

\end{psgopartialboard*}\par
\stone[\marklb{8}]{black}=\stone[\marktr]{white}\par
\tx ：白1扳好，黑2贪吃一子，不得要领。白3断后，再于5位``可立则立''，多送一子。至白11，两面均得包收之利，白十分生动。\par
\begin{psgopartialboard*}{(1,1)(8,8)}
	%11
	\stone{black}{c}{4}
	\stone{black}{d}{4}	
	\stone{black}{d}{6}
	\stone{black}{e}{5}
	\stone{black}{e}{6}
	\stone{black}{g}{3}
	
	\stone{white}{d}{5}
	\stone{white}{d}{8}
	\stone{white}{e}{3}
	\stone{white}{e}{4}
	\stone{white}{f}{5}
	\stone{white}{f}{6}
	\stone{white}{g}{4}
	\stone{white}{h}{3}
	
	\setcounter{gomove}{0}%步数编号从1开始
	\pass*
	\move{c}{6}%1
	\move{c}{5}%2
	\move{c}{7}%3
	\move{d}{3}%4

\end{psgopartialboard*}\par
\tx ：为了防止上图白3断、5立的手段，黑2提不得已，以下至黑4，是必然的结果。白1如果怕被吃一子，而在3位尖，被黑1位挡后，白大损，这样的下法，是不懂棋理的结果。\par
\begin{psgopartialboard*}{(1,1)(8,8)}
	%12
	\markpos{A}{f}{4}

	\stone{black}{c}{5}
	\stone{black}{c}{7}	
	\stone{black}{d}{4}
	\stone{black}{d}{5}
	\stone{black}{d}{6}

	\stone{white}{c}{6}
	\stone{white}{d}{2}
	\stone{white}{d}{7}
	\stone{white}{d}{8}
	\stone{white}{e}{6}
	\stone{white}{f}{3}

	\setcounter{gomove}{0}%步数编号从1开始
	\move{h}{3}%1

\end{psgopartialboard*}\par
\tx （习题）：这是古定式``双飞燕''产生的棋形，黑1夹后，就有A位压的手段，白应怎样走？\par
\begin{psgopartialboard*}{(1,1)(10,8)}
	%13	
	\markpos{A}{f}{4}
	
	\stone{black}{c}{5}
	\stone{black}{c}{7}	
	\stone{black}{d}{4}
	\stone{black}{d}{5}
	\stone{black}{d}{6}
	
	\stone{white}{c}{6}
	\stone{white}{d}{2}
	\stone{white}{d}{7}
	\stone{white}{d}{8}
	\stone{white}{e}{6}
	\stone{white}{f}{3}
	
	\setcounter{gomove}{0}%步数编号从1开始
	\move{h}{3}%1
	\move{b}{6}%2
	\move{b}{7}%3
	\move{b}{8}%4
	\move{b}{5}%5
	\move{c}{8}%6
	\move{a}{6}%7
	\move{h}{4}%8
	\move{j}{4}%9
	\move{g}{4}%10
	\move{k}{3}%11
	\move{c}{3}%12
	\move{b}{3}%13
	\move{b}{2}%14
	\move{a}{3}%15

\end{psgopartialboard*}\par
\begin{psgopartialboard*}{(1,1)(8,8)}
	%14
	\stone{black}{c}{5}
	\stone{black}{c}{7}	
	\stone{black}{d}{4}
	\stone{black}{d}{5}
	\stone{black}{d}{6}
	\stone{black}{h}{3}
	
	\stone{white}{b}{6}
	\stone{white}{c}{6}
	\stone{white}{d}{2}
	\stone{white}{d}{7}
	\stone{white}{d}{8}
	\stone{white}{e}{6}
	\stone{white}{f}{3}

	\setcounter{gomove}{0}%步数编号从1开始
	\move{f}{4}%1
	\move{g}{4}%2
	\move{g}{3}%3
	\move{e}{4}%4
	\move{f}{5}%5
	\move{e}{5}%6
	\move{e}{3}%7
	\move{d}{3}%8
	\move{f}{2}%9
	\move{b}{5}%10

\end{psgopartialboard*}\par
\tx ：白2立好手，这样虽多送一子，却产生了多种利用。至黑15的结果，白两面走到，棋形生动。白2立后，黑A压也不能成立，因为走成图十四的结果，黑不行。\par
\begin{psgopartialboard*}{(1,1)(9,7)}
	%15	
	\stone{black}{c}{4}
	\stone{black}{c}{5}	
	\stone{black}{c}{6}
	\stone{black}{e}{5}
	\stone{black}{e}{6}
	\stone{black}{e}{7}
	\stone{black}{h}{4}
	
	\stone{white}{b}{3}
	\stone{white}{c}{3}
	\stone{white}{d}{4}
	\stone{white}{d}{7}
	\stone{white}{e}{4}
	\stone{white}{f}{5}
	\stone{white}{f}{6}

	\setcounter{gomove}{0}%步数编号从1开始
	\move{f}{4}%1
	\move{g}{4}%2
	\move{f}{3}%3
	\move{g}{3}%4
	\move{e}{2}%5
	\move{e}{3}%6
	\move{f}{2}%7
	\move{g}{2}%8
	\move{d}{2}%9
	\move{d}{3}%10
	\move{g}{5}%11
	\move{j}{2}%12
	\move{j}{3}%13
	\move{h}{2}%14

\end{psgopartialboard*}
\begin{psgopartialboard*}{(1,1)(9,7)}
	%16
	\stone{black}{c}{4}
	\stone{black}{c}{5}	
	\stone{black}{c}{6}
	\stone{black}{e}{5}
	\stone{black}{e}{6}
	\stone{black}{e}{7}
	\stone{black}{h}{4}
	
	\stone{white}{b}{3}
	\stone{white}{c}{3}
	\stone{white}{d}{4}
	\stone{white}{d}{7}
	\stone{white}{e}{4}
	\stone{white}{f}{5}
	\stone{white}{f}{6}
	
	\setcounter{gomove}{0}%步数编号从1开始
	\pass*
	\move{f}{3}%1

\end{psgopartialboard*}\par
{\heiti 图十五}：这是小目高挂两间高夹定式的一型，黑1断、再3至9弃子，是符合``可立则立、能曲必曲''要诀的一法。本图和图十六白一手补尽的好形比较，希望能从中体会利用弃子，留有余味，在变化中得利的棋理，以及棋理和要诀的关系。\par
\begin{psgopartialboard*}{(1,1)(7,6)}
	%17
	\stone{black}{d}{2}
	\stone{black}{d}{3}	
	\stone{black}{e}{4}
	\stone{black}{f}{3}
	\stone{black}{f}{4}

	\stone{white}{d}{4}
	\stone{white}{d}{5}
	\stone{white}{e}{2}
	\stone{white}{e}{3}

	\setcounter{gomove}{0}%步数编号从1开始
	\pass*
	\move{f}{2}%1
	\move{g}{2}%2

\end{psgopartialboard*}
\begin{psgopartialboard*}{(1,1)(7,6)}
	%18
	\markpos{A}{g}{2}

	\stone{black}{b}{4}
	\stone{black}{c}{4}	
	\stone{black}{c}{6}
	\stone{black}{e}{4}
	\stone{black}{e}{5}
	\stone{black}{f}{3}

	\stone{white}{b}{3}
	\stone{white}{c}{3}
	\stone{white}{d}{3}
	\stone{white}{e}{3}
	\stone{white}{f}{4}
	\stone{white}{g}{3}

	\setcounter{gomove}{0}%步数编号从1开始
	\move{f}{2}%1
	\move{g}{4}%2

\end{psgopartialboard*}\par
{\heiti 图十七}：要诀中有``可''和``能''二字，说明是有条件的，必须加以辨别。本图白1曲不行。图十八黑1立也徒受损。\par
以上两图的结果说明：不能产生余味，不能在变化中得利，则不可``立''，不可``曲''，盲目弃子。\par
\clearpage
